¿^creative , ars mathematica ^commons contemporánea Also available at http://amc-journal.eu ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.) ARS MATHEMATICA CONTEMPORANEA 10 (2016) 67-77 Commutators of cycles in permutation groups Aleš Vavpetic * Fakulteta za Matematiko in Fiziko, Univerza v Ljubljani, Jadranska ulica 19, SI-1111 Ljubljana, Slovenija Received 10 January 2013, accepted 26 September 2014, published online 27 May 2015 Abstract We prove that for n > 5, every element of the alternating group An is a commutator of two cycles of An. Moreover we prove that for n > 2, a (2n + 1)-cycle of the permutation group S2n+1 is a commutator of a p-cycle and a q-cycle of S2n+1 if and only if the following three conditions are satisfied (i) n + 1 < p, q, (ii) 2n +1 > p, q, (iii) p + q > 3n +1. Keywords: Commutator, cycle, permutation, alternating group. Math. Subj. Class.: 20B05 1 Introduction In 1951 O. Ore [9] conjectured that in a finite simple non-abelian group every element is a commutator. In the same paper he proved that the conjecture holds for the alternating group An, where n > 5, but the result had already been proved by G. A. Miller half a century earlier [7]. After Ore published the paper there were many papers devoted to the Ore conjecture: R. C. Thompson proved the Ore conjecture for the projective special linear groups PSLn(q) [10], [11], [12], R. Gow proved it for the projective simplectic groups PSp2n(q), where q = 1 (mod 4) [4], O. Bonten for the exceptional groups of Lie type of low rank [2], J. Neubuser, H. Pahlings, E. Cleuvers proved it for the sporadic groups [8], E. W. Ellers, N. Gordeev handled the finite simple groups of Lie type over a finite field Fq, whenever q > 9, ... M. W. Liebeck, E. A. O'Brien, A. Shalev, P. H. Tiep proved the Ore conjecture for the remaining cases [6] and the conjecture became the theorem. We refer the reader to the survey paper [5] for more historical notes about commutators and the Ore conjecture. In this paper we prove a stronger version of the Ore conjecture for the simple alternating group An. In Section 2 it is shown that, for n > 5, every permutation of An is actually a *This research was supported by the Slovenian Research Agency grants P1-0292-0101 and J1-5435-0101. E-mail address: ales.vavpetic@fmf.uni-lj.si (Aleš Vavpetic) ©® This work is licensed under http://creativecommons.org/licenses/by/3.0/ 68 Ars Math. Contemp. 10 (2016) 31-44 commutator of two cycles of An. In particular, every even permutation of the symmetric group Sn is a product of two conjugate cycles. Namely, if p = [a, r] = a-1r-1 ar, then p is a product of a-1r-1a and r (and also a product of a-1 and r-1ar). Note that permutations r and r-1 are conjugate in Sn. In [1] it is proved that a (2n + 1)-cycle of A2n+1 is a product of two conjugate l-cycles of A2n+1 if and only if l > n +1. Hence this is a necessary condition for the existence of two l-cycles a and r such that [a, r] is a (2n + 1)-cycle. In Section 3 it is shown that this is far from being a sufficient condition. More precisely, it is shown that, for n > 2, a (2n + 1)-cycle of A2n+1 is a commutator of a p-cycle and a q-cycle of S2n+1 if and only if n +1 < p, q and p + q > 3n + 1. In particular, a (2n + 1)-cycle of A2n+1 (n > 2) is a commutator of l-cycles of S2n+1 if and only if l > 3n2ti. The image of an element a under a permutation a is denoted by aCT. Permutations are executed from left to right. The support supp a of a permutation a is the set of all elements which are not fixed by a. Let a be a permutation, a G supp a and x1,..., xn G supp a. We define permutations y(a; a, x1,..., xn) and e(a; a) by (X1, t = a, xi+1, t G {x1, . . . , xn-1}, a , t — xn, t , t G{a,x1,...,xn}, and t = a, = ^ aJ, t = aCT , ta, t G{a,aCT-1}. If a is the k-cycle (a1,..., ak), then y(a; ak, x1,..., xn) is the (k + n)-cycle (a1,..., ak, x1,..., xn) and e(a; ak) is the (k — 1)-cycle (a1,..., ak-1). Let a and r be permutations such that supp a n supp r = 0. For a G supp a and b G supp r, let -(a, r; a, b) denote the permutation defined by ,b) = ta, t G supp a — {a}, bT, t = a, tT, t G supp r — {b}, T, t = b. If r is a k-cycle then -(a, r; a, b) = y(a; a, bT, bT , bT ), and if a is a k-cycle then -(a, r; a, b) = y(r; b, aCT, a° ). 2 2 5) is a commutator of two cycles is based on induction on the number and the lengths of cycles in the cycle decomposition of the permutation. In the following lemmas we describe how the application of and e modify commutators. Ales Vavpetic: Commutators of cycles in permutation groups 69 Lemma 2.1. Let a, t be permutations, x G supp a, y G supp t, and (supp a U supp t) n ({xi,...,xn}U {yi,...,ym}) = 0. Then for t G {xT ,xTT, yTT, yT ' yi ,...,ym} we have ] = , 1, . . . , ^^n Proof. Denote a = ^>(a; x, x1,..., xn) and r = ^(t; y, y1,..., ym). For t G {xT, xTT, yTT ,yT 'tt ,xi,...,x„, yi,...,ym} we have tT 1 = tT \ Since t G {yTT ,yi,...,ym}, also tT G {yT, yi,...,ym} and therefore tT T = tT T . Since tT T G {x, xi,...,x„} we have tT_lT_l T = tT-lT-lTT. And finally tT_lT_l T G {y, yi,..., ym}, hence t^l = t^. □ We record the following immediate consequence. Corollary 2.2. Let a, t be permutations. Suppose that a, b G supp a such that aT = aT = b, and (supp a U supp t) n ({xi,..., xn} U {yi,..., ym}) = 0. Then for t G {bT, bTT, xi,..., xn, yi,..., ym} we have ttT'T] = t[v(T;b.Xi.....xn),v(T;6,yi,...,ym)]. Lemma 2.3. Let a, t be permutations and a, b G supp a such that b = aT = aT and c, d G supp a U supp t. Then [^(a; b, c, d) ^(t; b, d, c)] = ^Qa t]; bTT, c, d). Proof. Denote r = ^(a; b, c, d) and r = ^>(t; b, d, c). By Corollary 2.2, we have ttT'T] = t[T'T] for t G {bT, bTT, c, d}. Because i (bTCT)[([t, a]; bT,c, d)-i = = ^([a,T]; b1,d,c). □ Corollary 2.5. Let p be a (2n + 1)-cycle and n > 2. For p, q G N such that p, q < 2n +1 and p + q > 3n + 2, there exist a p-cycle a, a q-cycle t, and a G supp a such that [a, t] = p, supp p = supp a U supp t, and a1 = aT. In the case q = 2n +1 we arrange that a11 G supp t. Proof. If n = 2 and p > q then (p, q) G {(5, 5), (5,4), (5, 3), (4,4)} and we have (ai, a2, a3, a4, a5) = [(ai, a4, a2, a3, a5), (ai, a4, a3, a5, a2)] = = [(ai, a4, a2, a5, a3), (ai, a4, a3, a5)] = = [(ai, a2, a4, a5, a3), (ai, a2, a5)] = = [(ai,a5,a2,a3), (ai, a5, a3, a4)]. If n = 2 andp < q, then q = 2n +1 = 5 and we can use the equality [a, t]-i = [t, a]. In all cases ai = ai and if q = 5, also a^ G supp t. Let n > 2. The proof is divided into 3 cases. Case 1: Suppose q < 2n. Let pi = p - 2, qi = q - 1, and ni = n - 1. Then pi + qi = p - 2 + q - 1 > 3ni + 2 and pi, qi < 2ni + 1. By the inductive hypothesis there exist a pi-cycle a, a qi-cycle t, and a G supp a such that [a, t] is a (2ni + 1)-cycle, supp a U supp t = supp[a, t], and a1 = aT. Let x,y G supp a U supp t, T = ^(a; a1, x, y), and T = ^>(t; aT, y). Then a is a p-cycle, T is a q-cycle, a1 = a1 = aT = aT, a11 = x G supp T, and by Lemma 2.4, [T, T] is a (2n+1)-cycle and supp TUsupp T = supp[T, r]. Case 2: Suppose q = 2n +1 and p = 2n +1. This case follows from the previous case and equality [a, t]-i = [t, a]. Case 3: Suppose p = q = 2n + 1. By the inductive hypothesis there exist (2n - 1)-cycles a, t, and a G supp a such that [a, t] is a (2n - 1)-cycle, supp a = supp t = supp[a, t], and a1 = aT. Let x, y G supp a, T = ^(a; a1, x, y), and r = ^>(t; aT, y, x). Then T and r are (2n + 1)-cycles, aT = a1 = aT = aT, and by Lemma 2.3, [T, r] is a (2n + 1)-cycle and supp T = supp T = supp[T, r]. □ Lemma 2.6. Let a, t be permutations and a, b G supp a such that b = a1 = aT, b1 G supp t, and c G supp a U supp t. Then [a, ¥>(t ; b,c)]= £ ([a, t ]; b1 )(c,b1). Proof. Let T = ^(t; b, c). By Corollary 2.2, we get t[1'T] = t[1-T] for t G {b1, bT1, c}. From (b1 )[1'T] = bT-l1T = a1T = bT = c, c[1'T] = cT-l1T = b1T = b1, (bT1 )[1,T] = (bT )T-l1T = c1T = cT = bT, Ales Vavpetic: Commutators of cycles in permutation groups 71 and (bTCT )[ff'T 1 = baT = ba, (bCT )k>Tl = bT= it follows ; b, c)] = e([a,r]; bCT)(c,bCT). a bT, □ Corollary 2.7. Lei ni, n2 G N and let p be a product of two disjoint cycles of lengths 2ni and 2n2, respectively. If p, q < 2(ni + n2) — 1 andp + q > 3(ni + n2) then there exist a p-cycle a, a q-cycle t, and a G supp a such that p = [a, t], supp p = supp a U supp t, and aJ = aT. If ni = n2 = 1 then there exist no cycles a and t such that the length of one of them is strictly greater than 2(ni + n2) — 1 = 3, [a, t] is a product of two disjoint transpositions, supp[a, t] = supp a U supp t, where aJ = aT for some a G supp a. That means that in the Corollary in this case the upper bound requirement on the length of the cycles is sharp. If ni + n2 > 3 the upper bound requirement is not sharp (it can be increased to 2(ni + n2)) but the bound in the Corollary is in almost all cases sufficient for our purposes. Namely, in the case ni + n2 > 4, we get 2(2(ni + n2) — 2) > 3(ni + n2) and therefore the Corollary provides two cycles whose lengths can be required to be (independently) either odd or even: both odd (p = q = 2(ni + n2) — 1), both even (p = q = 2(ni + n2) — 2), the first even and the second odd (p = 2(ni + n2) — 2, q = 2(ni + n2) — 1), the first odd and the second even. Proof. One may assume that ni > n2. The proof is by induction on n2. Let n2 = 1. If ni = 1 then the only possibility for p and q is p = q = 3. In this case [(ai, a2, a3), (ai, a2, a4)] = (ai, a2), (a3, a4). Let ni > 2. Because p + (q — 1) > 3(ni + 1) — 1 = 3ni + 2 and p, q < 2(ni + 1) — 1 = 2ni + 1, Corollary 2.5 provides a p-cycle a, a (q — 1)-cycle t, and a G supp a such that [a, t] is a (2ni + 1)-cycle, supp a U supp t = supp[a, t], aJ = aT, and aJJ G supp t. Let c G supp a U supp t and r = ^(t; aT, c). Then r is a q-cycle, aJ = aT = aT, and by Lemma 2.6, [a, r] = e([a, t]; aJJ)(aJJ, c) and supp a U supp r = supp[a, r|. Note that aJTJ = c is in the support of the 2-cycle. For the proof by induction, suppose that for all n < n2 the assumptions p, q < 2(ni + n) — 1 and p + q > 3(ni + n) guarantee the existence of a p-cycle a, a q-cycle t, and a G supp a such that the following hold: [a, t] is a product of two disjoint cycles of lengths 2ni and 2n, supp[a, t] = supp a U supp t, aJ = aT, and aJTJ is in the support of the 2m-cycle in the cycle decomposition of [a, t]. We prove that the same holds for n = n2. The proof is divided into 3 cases. Case 1: Let q < 2(ni + n2) — 1. Definep = p — 2, q = q — 1, and m = n2 — 1. Because p + q > 3(ni + m) and pr, q < 2(ni + m) — 1, the inductive hypothesis yields a p-cycle a, a q-cycle t, and a G supp a such that [a, t] = pip2, where supp pi n supp p2 = 0, pi is a 2ni-cycle, p2 is a 2m-cycle, aJ = aT, and aJTJ G supp p2. Let x, y G supp a U supp t, r = ^(a; aJ,x,y), and r = ^(t; aT,y). Then r is a p-cycle, r is a q-cycle, aT = aJ = aT = aT, and by Lemma 2.4, [r, r] = ^(pip2; aJJ, x, y) = pi^(p2; aJJ, x, y) and aTTT = aJJ G supp ^(p2; aJJ,x,y). Case 2: Letp = 2(ni + n2) — 1 and q = 2(ni + n2) — 1. This case follows from the previous case and the equality [a, t]-i = [t, a]. 72 Ars Math. Contemp. 10 (2016) 31-44 Case 3: Letp = q = 2(n + n2) - 1. Definep = q = 2(n + n2) - 3 and m = n2 - 1. From pq, q < 2(n1 + m) - 1 and n1 > 1 we get p + ? > 3(n1 + m). By the inductive hypothesis there exist p-cycles a, t, and a G supp a such that [a, t] = p1p2, where supp p1 n supp p2 = 0, p1 is a 2n1-cycle, p2 is a 2m-cycle, aCT = aT, and aCTTCT G supp p2. Let x, y G supp a U supp t, 5 = <^>(a; aCT, x, y), and r = <^>(t; aT, y, x). Then 5 and r are p-cycles, aT = aa = aT = aT, and by Lemma 2.3, [5,?] = ^(p1p2; aCTTCT,x,y) = p1^(p2; aCTTCT, x, y) and aTTT = G supp ^(p2; aCTTCT, x, y). □ Lemma 2.8. Let a, t be permutations and a, b G supp a such that b = aCT = aT, and x, y, z G supp a U supp t. Then b(a; z),^(t ;z)] = [a,TKx,y, z). Proof. Let 5 = y(a; b, x, y, z) and r = ^>(t; b, y, z). By Corollary 2.2, we have t[T'T] = t[ff'tl for t G jbCT,bTff, x, y, z}. As (aCT )T = (bT)ffT = (bff )[ff'tl, |CTT = (bTCT )ict't ] (b^) T = zT lTT = yTT = zT = bT = (bTCT) T = (bT)T-1 TT _ z^T = bffT = x T = bT-:l = aTTT = bTT = y, y T = xt-1 = xTT = yT = z, z T = y^ = bTTT = xT = x, we have [r,r] = [a, t](x, y, z). □ Lemma 2.9. Let a1,a2,T1,T2 be cycles such that (suppa1 U suppt1) n (suppa2 U supp t2) = 0. Suppose there exist a G supp a1 and b G supp a2 such that aCTl = aT1 and bCT2 = bT2. Then [^(a1,a2; aCTl, bCT2),^(n,T2; aTl, bT2)] = [a^n][a2,T2]. Proof. Let a = ^(a1,a2; aCTl, bCT2) and t = ^(t1,t2; aTl, bT2). Set c = aCTl = aTl and d = bCT2 = bT2. From Corollary 2.2 and equalities a = ^(a1; c, b^2,..., b, bCT2) and t = ^(t1; c, bT22,... ,b, bT2), we get t^l = ttctl'tll = t[ctl'tll[ct2't2l for t G jcCTl ,cTlCTl} U supp a2 U supp t2. From Corollary 2.2 and equalities a = ^>(a2; d, aCTl,..., a, aCTl) and t = ^(t2; d, aTi,..., a, aTl), we get ttct't 1 = ttct2't21 = t[ffl-Tl]tct2't2] for t G jdCT2, dT2CTl }U suppa2 U suppt2. Therefore t^ = t[ctl'tll[ct2't2l for t G jcCTl, cTlCTl, dCT2, dT2CTl}. From (cffl )[ff'T] = dT-lffT = bffT = dT = cTl = afflTl = cTl-lfflTl = (cffl )iffl'tll, (cTlffl )[ff'T] = (cTl )T-lffT = dffT = (cffl )T = cfflTl = (cTlffl )iffl'tll, (dff2 )[ff'T] = cT-Vt = aCTT = cT = dT2 = bff2T2 = dT2-lff2T2 = (dff2 )[ff2't2l, (dT2ff2 )[ff'T] = (dT2 )T-2ffT = cffT = (dff2 )T = dff2T2 = (dT2 ff2 )[ff2't2l, weget [a, t] = [a1,T1][a2,T2]. □ Theorem 2.10. Let p G An. If n > 5 or p is not a 3-cycle then p is a commutator of two cycles of An. Ales Vavpetic: Commutators of cycles in permutation groups 73 Proof. If p = (ai,a2,a3) is a 3-cycle then n > 5 and p = [(ai,a3,x), (a;[,a2,y)] for some x, y G supp p. Suppose that p is not a 3-cycle. We show that there exist cycles a and t of odd lengths and a G supp a such that p = [a, t], supp p = supp a U supp t, and aCT = aT. The proof is by induction on the number of cycles in the cycle decomposition of p, which we denote by c(p). If c(p) = 1, p is a cycle of odd length l > 5. The statement follows from Corollary 2.5. If c(p) = 2, then let p = p1p2, where p1 and p2 are disjoint cycles. The lengths of these cycles are of the same parity. If the lengths are even, the statement follows from Corollary 2.7. In the case of odd lengths, 3 cases are considered. Case 1: Suppose both lengths are 3. Then [(a1, a2, a6, a5, a3), (a1, a2, a4, a6, a5)] = (ai, a2, a3)(a4, a5, a6). Case 2: Suppose exactly one of the lengths is 3. One may assume p2 = (x, y, z) is the 3-cycle. Let p1 be a cycle of length 2l + 1, where l > 2. By Corollary 2.5, there exist a 2l-cycle a, a (2l + 1)-cycle t, and a G supp a such that p1 = [a, t], supp p1 = supp a U suppt, and aF = aT. By Lemma 2.8, we have p = [y(a; aCT, x, y, z),y(T; aT, y, z)], where ^(a; aCT, x, y, z) and ^(t; aT, y, z) are (2l + 3)-cycles. Case 3: Suppose both lengths are greater than 3. Let pi be a cycle of length 2lj + 1, li > 2. By Corollary 2.5, there exist (2l1 + 1)-cycles a1, t1, (2l2)-cycles a2, t2, a1 G suppa1, and a2 G suppa2 such that pi = [ai,Ti], supppi = suppai U suppTi, and a1i = aTii. Then ^(a1, a2; a^1, aa22) and ^(t1, t2; a^1, a^2) are (2(l1 +12) + 1)-cycles and by Lemma 2.9, p = [^(a1, a2; a^1, aip), ^(t1, t2; a^1, a^2)]. If c(p) > 3, the following 4 cases are considered. Case 1: Suppose p = p1p2, where p2 is a (2l+1)-cycle, l > 2, and supp p1 nsupp p2 = 0. By Corollary 2.5, there exist (2l)-cycles a2, t2 and b G suppa2, such that p2 = [a2, t2], suppp2 = suppa2 U suppt2, and bCT2 = bT2. Because 2 < c(p1) < c(p) - 1, the inductive hypothesis yields cycles a1, t1 of odd lengths, as well as a G supp a1, such that p1 = [a1, t1], supp p1 = supp a1 U supp t1, and aCT1 = aT1. By Lemma 2.9, we have p = [^(a1,a2; aCT1, bCT2),^(t1,t2; aT1, bT2)], where ^(a1,a2; aCT1, bCT2) and ^(t1,t2; aT1, bT2) are cycles of odd lengths. Case 2: Suppose p = p1p2, where p2 = (a1, a2, a3)(a4, a5, a6) and supp p1 n supp p2 = 0. If p1 = (a7,a8,ag) then p = [(a1, a2, a7, as, ag, a4, a5, a3, a6), (a1, a2, as, ag, a5, a3, a4)]. If p1 is not a 3-cycle, the inductive hypothesis yields cycles a1, t1 of odd lengths, as well as a G supp a1, such that p1 = [a1,T1], supp p1 = supp a1 U supp t1, and aCT1 = aT1 = b. Then a = ^(^(a1; b, a1, a2, a3); b, a4, a5, a6) and t = y(y(T1; b, a2, a3); b, a5, a6) are cycles of odd lengths and, using Lemma 2.8 twice, we get p = [a, t]. Case 3: Suppose p = p1p2, where p2 is a disjoint product of cycles of lengths 2l1 and 2l2, such that l1 + l2 > 3, and supp p1 n supp p2 = 0. If p1 = (a1, a2, a3) then by Corollary 2.7, there exist a (2(l1 + l2) - 2)-cycle a2, a (2(l1 + l2) — 1)-cycle t2, and a G suppa2, such that p2 = [a2, t2], suppp2 = suppa2 U suppt2, and aCT2 = aT2 = b. Then a = ^>(a2; b, a1, a2, a3) and t = ^(t2; b, a2, a3) are (2(l1 + l2) + 1)-cycles and by Lemma 2.8, we get p = [a, t]. If p1 is not a 3-cycle then by the inductive hypothesis there exist cycles a1, t1 of odd lengths and a G supp a1, such that p1 = [a1, t1], supp p1 = supp a1 Usupp t1, and aCT1 = aT1 .If l1 + l2 =3 then p2 = (a1, a2, a3, a4)(a5, a6) and for a2 = (a1, a5, a2, a4, a6, a3) and t2 = (a1, a5, a3, a4) we get p2 = [a2,T2] and for b = a1 we get bCT1 = bT1. If l1 +12 > 3 Corollary 2.7 provides (2(l1 +12) — 2)-cycles a2 andt2, as well as b G suppa2, such that 74 Ars Math. Contemp. 10 (2016) 31-44 P2 = [a2, t2], suppp2 = suppa2 U suppt2, and b72 = bT2. Then a = ^(ai, a2; aCTl, b72) and t = ^ (t1 , t2 ; aTl, bT2) are cycles of odd length and by Lemma 2.9, we get p = [a, t]. Case 4: Suppose p is a disjoint product of transpositions and at most one 3-cycle. If there are at most four transpositions in the cycle decomposition of p we have 3 possibilities: [(01,03, 05, 06, a2, 04, 07), (01, 03, 06, 04, 07, 02, 05)] = (01, 02x03, 04x05, 06, 07), [(ai,a2, 04, 08, 06, 03, 05), (01, 02, 03, 08, 04, 06, 07)] = (01, 02X03, 04)(05, 06)(07, 08), [(01,02, 05, 03, 04, 09, 010, 07, 011), (01, 02, 06, 03, 04, 010, 07, 09, 08)] = = (01, 02X03, 04X05, 06)(07, 08)(09,010, 011). Otherwise p = p1p2, where p2 = (01,02)(03,04)(05,06)(07, 08), 2 < c(p1) < c(p), and supp p1 n supp p2 = 0. By the inductive hypothesis there exist cycles a1, t1 of odd lengths and 0 G suppa1, such that p1 = [a1,T1], suppp1 = suppa1 U suppt1, and 07 = 0Tl. For a2 = (01,08,03,02,04,06,07,05) and T2 = (01,08,04,03,05,06) we have p2 = [a2, t2]. Then a = ^(a1, a2; 0CTl, 0^2) and t = ^(t1, t2; 0Tl, 0i2) are cycles of odd lengths and by Lemma 2.9, we get p = [a, t]. □ 3 Cycles as commutators of cycles From the previous section we know that a (2n + 1)-cycle is a commutator of a p-cycle and a q-cycle if p + q > 3n + 2 (and p, q < 2n + 1). But this sufficient condition is not necessary. Note that in the previous section we were interested in pairs of cycles a and t, for which there exists 0 G supp a such that 07 = 0T. We needed that for "concatenation" of cycles in Lemma 2.9. With that assumption withdrawn, the result is obtained by using a more stringent hypothesis as shown in the next corollary. Lemma 3.1. Let a, t be permutations, x, y G supp a U supp t, 01, 02 G supp a n supp t, b G supp a — supp t, and c G supp t — supp a, such that 07 = b, b7 = 02, 0i = c, and cT = 02. Then [^(a; ^ c, x) ^(t; c y)] = yQa t]; c, y, x). Proof. Let a = ^(a; b, c, x) and r = ^(t; c, y). If t G {x, 02,c} then t7 l = t7 l. If t G {y,02} then t7- G {y,02} and t7-lT-l = t7"l7_l. If t G {x, 07,02} then t7-lT-l G {x,c, b} and t7-lT-l7 = t7-T-l7. If t G {y^f } then t7-T-l7 G {y,c} and t7-lT-l7T = t7-lT-l77. Hence for t G {x,y,c, 02,07} we get t[7'T] = t[7'7]. Because c[7 T] = cT - l7T = 07T = bT = b, c[7 T7] = b7- l 77T7 = b77 = cT7 = y, y[7 T7] = y7- l 77T7 = c77 = xT7 = x, x[7 T7] = c7- l 77T7 = 0777 = bT7 = b, [7 02 T7] = x7- l 77T7 = x77T7 = 0T27 T = 0T2 = b7T = bT l7T = 027't], (07 )[7 T7] = °2 l 77T7 = y77 = y7 = 02 = cT = c7T = 02-l7t = (07 )[7'T we get [a, r] = y([a, t]; c, y, x). □ Ales Vavpetic: Commutators of cycles in permutation groups 75 Corollary 3.2. Let p be a (2n + 1)-cycle and n > 2. For p, q G N such that p, q < 2n and p + q = 3n +1, there exist a p-cycle a and a q-cycle t, such that [a, t] = p and supp p = supp a U supp t. Proof. By induction on n we prove that whenever p, q < 2n and p + q = 3n +1, there exist a p-cycle a, a q-cycle t, ai , a2 G supp a n supp t, b G supp a — supp t, and c G supp t — supp a, such that a4 = b, b4 = a2, ai = c, cT = a2, [a, t] is a (2n + 1)-cycle, and supp[a, t] = supp a U supp t . Because [t, a] = [a, t]-1 we may assume p > q. If n = 2 then p = 4, q = 3 and we have [(a1, b, a2, d), (a1, c, a2)] = (a1, c, b, d, a2). Let n > 2. For p, q < 2n and p + q = 3n +1 we define p = p — 2 and f = q — 1. Then pi + f = 3(n — 1) + 1 and pi < 2(n — 1). From q < p we get q = 2n and therefore f < 2(n — 1). By the inductive hypothesis there exist a pi-cycle i, a f-cycle i, a1, a2 G supp insupp i, b G supp i — supp f, and c G supp f—supp i, such that aT = b, b4 = a2, aT = c, cT = a2, [i, r] is a (2n — 1)-cycle, and supp[i, r] = suppi U suppi. Let x, y G supp f U supp f. Then a = y(i; b, c, x) is a p-cycle, t = y(f; c, y) is a q-cycle, c, a2 G supp a n supp t, x G supp a — supp t, y G supp t — supp a, c4 = x, x4 = a2, cT = y, yT = a2, and by Lemma 3.1, [a, t] is a (2n + 1)-cycle. □ Let a and t be permutations. An equivalence relation on the set supp a n supp t is defined in the following way. Elements a, b G supp a n supp t are equivalent if and only if there exist a0,..., an G supp a n supp t and p1,..., pn G {a, a-1, t, t-1}, such that a = a0, b = an, and = a?- 1 for i = 1,..., n. This is obviously an equivalence relation. Definition 3.3. Permutations a and t are braided if all elements of supp a n supp t are equivalent to each other. Lemma 3.4. Let a and t be cycles such that the commutator [a, t] is a cycle and supp[a, t] = supp a U supp t. Then a and t are braided. Proof. Let p = [a, t] and a0 G supp a n supp t . For n > 0 we inductively define a4n+1 = a4n , a4n+2 = a4n+3 = a4n+2, and a4n+4 = a4n+3. Let us show that if a4m = ap G supp a n supp t, then a4m is equivalent to a0. Let b1 = a0 and ii = max{i | i < 4m, aj = a0}. For k > 1 and < 4m we let ik+1 = max{i | ik < i < 4m, aj = aifc+i}, bfc+i = aifc+1, and pk G {a, a-1 ,t,t-1}, where pk is uniquely defined by bpkk = bk+1. If we show that bk G supp a n supp t for all k, then by definition, a0 = b1 is equivalent to a4m = b;. For 1 < k < l we have bk+1 G supp pk. Suppose bk+1 G supp f, where f is the cycle in {a, t} — {pk,p-1}. Because apk = aik+1 and pk = f±1, necessarily also a4k+1 = aifc+2 or aTrfc+11 = aifc +2. Because apfcfc+2 = aifc +3 and aik+1 G supp i, we get aik = aik+3. This contradicts the definition of ik. Hence bk G supp a n supp t. Let b G supp a n supp t. Because p is a cycle and b G supp p, there exists m such that b = a0 . Thus b is equivalent to a0, and hence a and t are braided. □ Lemma 3.5. Let a and t be permutations such that supp[a, t] = supp a U supp t. Then | supp a — supp t |, | supp t — supp a | < | supp a n supp t |. Proof. Suppose there exist x, y G supp a — supp t, such that x = y4. Then xt4't] = x, and consequently x G supp[a, t], which is a contradiction. Hence the map (supp a — 76 Ars Math. Contemp. 10 (2016) 31-44 supp t) ^ (supp a n supp t), defined by x ^ xa, is an injection. Therefore | supp a — supp t | < | supp a n supp t |. Because supp[T, a] = supp[a, t], the other inequality follows from the above paragraph. □ Lemma 3.6. Let a and t be cycles such that [a, t] is a cycle and supp[a, t] = supp a U supp t. Then | supp a — supp t | + | supp t — supp a| < | supp a n supp t | + 1. Proof. Let k = | supp a n supp t| supp a| = k + p, and | supp t| = k + q. If p = 0, then by Lemma 3.5 we have | supp a — supp t | + | supp t — supp a | = | supp t — supp a | < | supp a n supp t | + 1. Analogously for q = 0. Let p, q > 0. Let supp a — supp t = jai,..., ap}. Let m; G N U {0} be the largest number such that a^3 G supp a n supp t for all j G {1,..., m;}. We claim that all m; are positive. Indeed, suppose that there exist x, y G supp a — supp t, such that xa = y. Then ytct't] = y which is a contradiction since supp a c supp[a, t]. Hence the set M; = ja^,..., } is nonempty for all i. Because a is a cycle and p > 0, for every x G supp a n supp t there exists the smallest i G N such that xa * = ak for some k, which means that x G Mk. Therefore, (supp a n supp t) = M1 ]J ... ^ Mp. Similarly, (supp a n supp t) = N1 ]J ... ^ Nq, where supp t — supp a = jb1,..., bq}, N; = jbT,..., bT} C supp t n supp a, and bTG supp a. By Lemma 3.4, the cycles a and t are braided. Hence there exist i2 G {2,... ,p}, d2 G M1, c2 G Mi2, and t2 g {t, t-1} such that d2 = c222. For j > 2 there exist j G {2,... ,p} — ji2,..., ij-1}, d, G M1 U (ujr21Mii), c, G Mi3, and t, g {t, t-1} such that dj = j. Let us show that for each i, the set N; = N; — {c2,..., cp} is nonempty. By construction, the elements c2,..., cp are different, d, = ck for j < k, and every pair {c,, dj} is a subset of N; for some l. Suppose N; n {c2,..., cp} = {ckl,..., ckr}, where k1 < ... < kr. Then dfcl G N; and dfcl G {ck1,..., cfcr}, so dfcl G NVj = 0. Hence in the union of the q nonempty sets NV1,..., Nq there are exactly k — (p — 1) elements. This means that | supp t—supp a| = q < k — (p— 1) = | supp a nsupp t | — | supp a—supp t | + 1. □ Theorem 3.7. Lei n > 2 and lei p be a (2n + 1)-cycle. There exist a p-cycle a and a q-cycle t such that p = [a, t] and supp p = supp a U supp t if and only if the following three conditions are satisfied (i) n +1 < p, q, (ii) 2n + 1 > p, q, (iii) p + q > 3n +1. Proof. Suppose there exist a p-cycle a and a q-cycle t such that p = [a, t] and supp p = supp a U supp t. Let k = | supp a n supp t |, p = k + V, and q = k + V. By Lemma 3.5, we have V < k, therefore 2V < k + V = q < 2n +1 which implies V < n. Then 2n +1 = | supp p| = | supp a U supp t | = p + V < p + n, hence n +1 < p. By Lemma 3.6, we have p + V < k +1. Therefore 2n +1 = k + pV + V < 2k + 1 and p + q = 2n +1 + k > 3n +1. If p + q > 3n + 2 the theorem follows from Corollary 2.5. If p + q = 3n +1, the theorem follows from Corollary 3.2. □ References [1] E. Bertram, Even permutations as a product of two conjugate cycles, J. Combin. Theory 12 (1972), 368-380. Ales Vavpetic: Commutators of cycles in permutation groups 77 [2] O. Bonten, Über Kommutatoren in endlichen einfachen Gruppen, Aachener Beiträge zur Math. 7, Verlag der Augustinus-Buchhandlung, Aachen, 1993. [3] E. W. Ellers, N. Gordeev, On the conjectures of J. Thompson and O. Ore, Trans. Amer. Math. Soc. 350 (1998), no. 9, 3657-3671. [4] R. Gow, Commutators in the symplectic group, Arch. Math. (Basel) 50 (1988), no. 3, 204-209. [5] L. C. Kappe, R. S. Morse, On commutators in groups, Groups St. Andrews 2005. Vol. 2, 531558, London Math. Soc. 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